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For, if it be not, through A draw AG parallel to BF, (1. 31) and join GF.

DEMONSTRATION

Because the triangles ABC, GEF, are upon equal bases BC, EF, and are assumed to be between the same parallels BF, AG;

therefore the triangle ABC must be equal to the triangle GEF; (1. 38) but the triangle ABC is equal to the triangle DEF, (hyp.)

therefore also the triangle DEF must be equal to the triangle GEF, (ax. 1)

the greater to the less, which is impossible;

therefore AG is not parallel to BF.

And in the same manner it can be demonstrated that there is no other parallel to it but AD;

therefore AD is parallel to BF.

Wherefore equal triangles, &c.

Q. E. D.

PROP. XLI.— THEOREM.

If a parallelogram and a triangle be upon the same base and between the same parallels; the parallelogram shall be double of the triangle.

(References-Prop. 1. 34, 37; ax. 1.)

Let the parallelogram ABCD, and the triangle EBC be upon the same base BC, and between the same parallels BC, AE.

Then the parallelogram ABCD shall be double of the triangle EBC.

D

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Because the triangles ABC, ECB are upon the same base BC, and between the same parallels BC, AE,

therefore the triangle ABC is equal to the triangle EBC; (1. 37) but the diameter AC bisects the parallelogram ABCD,

therefore the parallelogram ABCD is double of the triangle ABC, (1. 34) wherefore also‘ABCD is double of the triangle EBC. (ax. 1) Therefore if a parallelogram, &c.

Q. E. D.

PROP. XLII.— PROBLEM.

To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

(References

Prop. 1. 10, 23, 31, 38, 41; ax. 6; def. 1. 34.)

Let ABC be the given triangle, and D the given rectilineal angle. It is required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D.

A F G

CONSTRUCTION

Bisect BC in E, (I. 10) join AE,

and at the point E in the straight line EC, make the angle CEF equal to D; (I. 23)

through A draw AFG parallel to EC, and through C draw CG parallel to EF. (I. 31.)

Then FECG is the parallelogram required.

DEMONSTRATION

Because the base BE is equal to the base EC, and BC parallel to AG, therefore the triangle ABE is equal to the triangle AEC; (1. 38) and therefore the triangle ABC is double of the triangle AEC; but because the parallelogram FECG and the triangle AEC are on the same base EC, and between the same parallels EC, AG,

therefore the parallelogram FECG is double of the triangle AEC; (1. 41)

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but things which are double of the same are equal to one another' (ax. 6)

therefore the parallelogram FECG is equal to the triangle ABC,

and it has one of its angles CEF equal to the given angle D. (constr.)

Wherefore there has been described a parallelogram FECG equal to a given triangle ABC, having one of its angles CEF equal to the given angle D. Q. E. F.

PROP. XLIII.—THEOREM.

The complements of the parallelograms which are about the diameter of any parallelogram, are equal to one another.

(References Prop. I. 34; ax. 2, 3.)

Let ABCD be a parallelogram, of which the diameter is AC; and EH, FG, the parallelograms about AC, that is, through which AC passes;

and let BK, KD be the other parallelograms, which make up the whole figure ABCD, and are therefore called the complements.

Then the complement BK shall be equal to the complement KD.

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Because ABCD is a parallelogram, and AC its diameter,

therefore the triangle ABC is equal to the triangle ADC; (1. 34) and because EKHA is a parallelogram, and AK its diameter,

therefore the triangle AEK is equal to the triangle AHK;

and for the same reason,

the triangle KGC is equal to the triangle KFC.

And, because the triangle AEK is equal to the triangle AHK, and the triangle KGC to KFC;

therefore the two triangles AEK, KGC, are equal to the two triangles, AHK, KFC; (ax. 2)

but the whole triangle ABC is equal to the whole triangle ADC;

therefore the remaining complement BK is equal to the remaining complement KD. (ax. 3.)

Wherefore, the complements, &c.

Q. E. D.

PROP. XLIV.- PROBLEM.

To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

(References - Prop. I. 15, 29, 31, 42, 43; ax. 9. 12.)

Let AB be the given straight line, and C the given triangle, and D the given rectilineal angle.

It is required to apply to the straight line AB, a parallelogram equal to the triangle C, and having an angle equal to D.

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Make the parallelogram BEFG equal to the triangle C, (1. 42)

and having the angle EBG equal to the angle D,

so that BE be in the same straight line with AB; produce FG to H,

and through A draw AH parallel to BG or EF, (1. 31) and join HB.

Then because the straight line HF falls upon the parallels AH, EF, therefore the angles AHF, HFE, are together equal to the two right angles; (I. 29)

wherefore the angles BHF, HFE, are less than two right angles; (ax. 9) but 'straight lines which with another straight line make the interior angles upon the same side less than two right angles, do meet if produced far enough;' (ax. 12)

therefore HB, FE, shall meet, if produced;

let them meet in K,

and through K draw KL parallel to EA or FH, (1. 31)

and produce HA, GB, to the points L, M.

Then LB shall be the parallelogram required.

DEMONSTRATION

Because HLKF is a parallelogram, of which the diameter is HK, and AG, ME, are the parallelograms about HK,

also LB, BF, are the complements;

therefore the complement LB is equal to the complement BF; (1. 43) but the complement BF is equal to the triangle C; (constr.)

wherefore LB is equal to the triangle C.

And because the angle GBE is equal to the angle ABM, (1. 15) and likewise to the angle D,

therefore the angle ABM is equal to the angle D. Wherefore the parallelogram LB is applied to the straight line AB, and is equal to the triangle C, and has the angle ABM equal to the angle D. Q. E. F.

PROP. XLV.-PROBLEM.

To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle.

(References Prop. I. 14, 29, 30, 42, 44; ax. 1, 2; 1. 34, def.)

Let ABCD be the given rectilineal figure, and E the given rectilineal angle.

It is required to describe a parallelogram equal to ABCD, and having an angle equal to E.

CONSTRUCTION

Join AC, and describe the parallelogram FH equal to the triangle ABC, and having the angle HKF equal to the angle E; (1. 42) and to the straight line GH apply the parallelogram GM equal to the triangle ADC, having the angle GHM equal to the angle E. (1. 44.) Then the figure FKML shall be the parallelogram required.

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